Summary_of_U_Substitution

Summary_of_U_Substitution - 1 9 u e du Now we can integrate...

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Summary of U-Substitution 1. ( 29 8 3 4 7 x x dx + Always let u equal the thing in the parentheses raised to the power. 4 7 u x = + Then take the derivative of u . 3 4 du x dx = Now we have to substitute back into the integral, but we don’t have a 3 4 x dx in our Problem so we need to get rid of the 4. So… 3 1 4 du x dx = Now we can substitute back into the integral 8 1 4 u du Now we can integrate (add one to the power and divide by it) 9 9 1 1 4 9 36 u C u C + = + But we start in terms of x so we must end in terms of x . ( 29 9 4 1 7 36 x C + + DONE! 2. 3 2 3 x x e dx
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Always let u equal the exponent of the exponential. 3 3 u x = Then take the derivative of u . 2 9 du x dx = Now we have to substitute back into the integral, but we don’t have a 2 9 x dx in our Problem so we need to get rid of the 9. So… 2 1 9 du x dx = Now we can substitute back into the integral
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Unformatted text preview: 1 9 u e du Now we can integrate (the anti-derivative of u e is u e ) 1 9 u e C + But we start in terms of x so we must end in terms of x . 3 3 1 9 x e C + DONE! 3. 1 6 7 dx x + Always let u equal the denominator. So 6 7 u x = + . 6 7 u x = + Then take the derivative of u . 6 du dx = Now we have to substitute back into the integral, but we dont have a 6 dx in our Problem so we need to get rid of the 6. So 1 6 du dx = Now we can substitute back into the integral 1 1 6 du u Now we can integrate (the anti-derivative of 1/u is ln(u)) 1 ln 6 u C + But we start in terms of x so we must end in terms of x . ( 29 1 ln 6 7 6 x C + + DONE!...
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Summary_of_U_Substitution - 1 9 u e du Now we can integrate...

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